If using voltage divider equation: Vadj = Vout x (R2/(R1+R2)) right? (the symbols used are wrt the circuit on the link, and Vadj is 1.25)
So if I rearrange, shouldn't I get Vout = 1.25 x (1 + R1/R2)??

The LM317 itself it not connected to ground.
The Vout is regulated to 1.25V higher than Vadj.
The output voltage is set higher than 1.25V by lifting the Vadj above ground level.

The voltage over R1 is always 1.25V. The current through R1 is 1.25/R1.
That same current has to go through R2. Also the current of the LM317 itself is going through R2, but that is only 50uA, so that is not in the formula.

If R1 = 100 ohm, and R2 = 200 ohm. The voltage over R1 is 1.25V and over R2 is 2.50V. The output will be 1.25+2.50 = 3.75V

Erdin:
The LM317 itself it not connected to ground.
The Vout is regulated to 1.25V higher than Vadj.
The output voltage is set higher than 1.25V by lifting the Vadj above ground level.

The voltage over R1 is always 1.25V. The current through R1 is 1.25/R1.
That same current has to go through R2. Also the current of the LM317 itself is going through R2, but that is only 50uA, so that is not in the formula.

If R1 = 100 ohm, and R2 = 200 ohm. The voltage over R1 is 1.25V and over R2 is 2.50V. The output will be 1.25+2.50 = 3.75V

So it does not use the voltage divider rule in this case?

And the inside of the LM317 is a non- inverting op amp right?
Similar to this: noninverting op-amp gain calculator
Where Rg is R2, Rf is R1 and Vin is the reference voltage of 1.25V.
Then what is the Voltage at V-? Is it 1.25V also? If so, why?

Your guess would have been right if the LM317 referenced the 1.25V to the ground terminal - but it doesn't
it references it to the output terminal (the technical reason for this is to provide a known closed-loop gain
throughout its range of operation). You apply the voltage divider rule upside-down, that's all.

You may want to look up operational amplifier gain. The gain in the simplest case is the ratio of two resistors. The assumption is only that the open loop gain (ie without resistors) is very large.

Can some kind soul explain to me how this circuit works?
As in how are the voltages at the inverting and non-inverting input determined and stuff

Thanks in advance

Well you should be confused since they got the formula wrong on the page for starters.

As for the schematic, when you tie the output of an op-amp directly back to the inverting (-) input, you create a "voltage follower". The op-amp will strive to the keep the output voltage at a level the same as the voltage at the non-inverting (+) input of the op-amp. The transistor that the op-amp output is going thru increases the current output greatly over what the op-amp itself can supply. The non-inverting (+) input of the op-amp is biased at 1.25V because of the Zener diode. The bias increases as R2 goes up. This drives the output higher in voltage.

The output of the op-amp will actually be .6V higher than the voltage being fed back which is the same as the output voltage of the transistor.

@don, Vin is not part of the math for calculating Vout.